ACIDS

What is acid?

An acid is a chemical substance, which ionizes in water to produce hydrogen ions, H+

When acids dissolved in water, the hydrogen atoms in acids are released as hydrogen ions, H+.

The hydrogen ion combined with water molecule, H2O to form a hydroxonium ion, H3O+. However, this ion can be written as H+.

We can classify an acid as a monoprotic acid or a diprotic acid based on basicity.

Basicity of an acid is the number of mole of H+ ion that can be produced by one mole of acid when it dissolves in water.

Monoprotic acidIonisation

Hydrochloric acid, HCl

Ethanoic acid, CH3COOH

Diprotic acidIonisation

Sulphuric acid, H2SO4

BASESWhat is base?

A base is a chemical substance that can neutralize an acid to produce a salt and water.

Examples of bases are;

i. metal hydroxidesii. metal oxide

Most bases are not soluble in water. Bases that are soluble in water are known as alkalis.

Bases that are insoluble in waterBases that are soluble in water

Zinc oxide, ZnOSodium oxide, Na2O

Zinc hydroxide, Zn(OH)2Sodium hydroxide, NaOH

Copper(II) oxide, CuO2Potassium oxide, K2O

Copper(II) hydroxide, Cu(OH)2 Potassium hydroxide, KOH

Calcium hydroxide, Ca(OH)2

Ammonia, NH3

Role of water and the properties of acids

Results

AcidConditionObservationInference

Ethanoic acid(molecule)Glacial (dry)- No colour change in the litmus paper- Light bulb is not lighted upDoes not show any acidic properties

Aqueous - Blue litmus paper is changed to red- Light bulb is lighted upShows acidic properties

Hydrogen chloride(molecule)Dissolved in methylbenzene- No colour change in the litmus paper

- Light bulb is not lighted upDoes not show any acidic properties

Aqueous- Blue litmus paper is

changed to red

- Light bulb is lighted upShows acidic properties

Discussion

The hydrogen ions, H+ are responsible for acids to show their properties1. Glacial/dry ethanoic acid does not show acidic properties because without water, ethanoic acid

remains or exists as molecules and there are no hydrogen ions present. 2. Aqueous ethanoic acid shows acidic properties because in water, ethanoic acid ionises to form hydrogen ions, H+.4. Hydrogen chloride gas in methylbenzene does not show acidic properties and does not conduct

electricity, because it exists as covalent molecules.

5. Hydrogen chloride gas in water shows acidic properties and conduct electricity. This is because

hydrogen chloride ionises in water to form hydrogen ions, H+ thus hydrochloric acid is formed.6. Aqueous ethanoic acid and hydrochloric acid can conduct electricity because in the solutions the ions can move freely. Conclusion

Chemicals Properties of Acid

a. Acids react with bases to form salts and water. Examples;

i. H2SO4 + ZnO

ii. 2HCl + CuO

b. Acids react with alkalis to form salts and water. Examples;

i. HCl + NaOH

ii. H2SO4 + Ca(OH)2

c. Acids react with reactive metal to produce salts and hydrogen gas.

Examples;

i. 2HCl + Mg

ii. H2SO4 + Zn

Reactive metals: Mg, Al, Zn (use this metals only)

d. Acids react with carbonate compound to produce salts, water and carbon

dioxide.

Examples;

i. 2HCl + PbCO3

ii. H2SO4 + CuCO3 Role of water and the properties of alkalis

Results

AcidConditionObservationInference

Ammonia

(molecule)In tetra chloromethane- No colour change in

the litmus paper

- Light bulb is not lighted upDoes not show any alkalis properties

Aqueous - Blue litmus paper is

changed to red

- Light bulb is lighted upShows alkalis properties

Discussion

The hydroxide ions, OH- are responsible for alkalis to show their properties1. Ammonia liquid in tetrachloromethane does not show alkalis properties because it exists as molecules.2. Aqueous ammonia acid shows acidic properties because in water, ethanoic acid ionises to form hydroxide ions, OH-.

Ionisation equation for ammonia;

3. Hydroxide ion, OH- in ammonia solution / solution hydroxide cause solution shows alkalis properties.

4. Ammonia gas in methylbenzene does not show alkalis properties and does not conduct electricity because it exits as a molecules.

5. Ammonia gas in water shows alkalis properties and conduct electricity because in the solutions the ions can move freely.

Conclusion

Chemicals Properties of Alkalisa. Alkalis react with acids to form salts and water.

Examples;

i. NaOH + HCl

ii. Ca(OH)2 + H2SO4 b. When a mixture of an alkali and an ammonium salt is heated, ammonia gas is, NH3 is liberated.

Examples;

i. NaOH + NH4Cl

ii. Ca(OH)2 + 2NH4Cl The Strength of Acids and Alkalis

1. The pH scale us used to indicate the degree of acidity or alkalinity of a solutions.

Strong and Weak Acids/Alkalis1. The strength of an acid or alkali depends on the degree of ionisation or dissociation of the acid or alkali in water.

2. Strong Acid: An acid which ionises completely in water. Example; HCl

(monoprotic acid)

H2SO4

(diprotic acid)

3. Weak Acid: An acid which ionises partially in water.

Example; CH3COOH

4. Strong Alkali: An alkali which ionises completely in water. Example; NaOH 5. Weak Alkali: An alkali which ionises partially in water.

Example; NH3 + H2O NH4 + + OH-

Concentrations of Acids and Alkalis

1. Concentration of solution can be expressed in gdm-3 or mol dm-3.

Example. 1:

A student dissolves 50.0 g of anhydrous copper(II) sulphate in water to make a 250 cm3 of solution.

What is the concentration of the solution in mol dm(3?

Solution:Mass of anhydrous copper(II) sulphate, CuSO4 = 50.0 g

Volume of solution = = 0.25 dm3

( Concentration of CuSO4 solution = = 200.0 g dm(3 200 gdm-3 = 200 mol dm-3

64 + 32 + 4x16

molar mass

Example 2:

28.0 g of potassium hydroxide is dissolved in water to make 200 cm3 of solution.

Calculate the molarity of potassium hydroxide solution obtained. (Ar: H, 1; O, 16; K, 39) Ans: 2.5 mol dm(3Solution:

Question 1

Find (a) the concentration of a sodium in grams per dm3 when 36.5 g of hydrogen chloride, HCl is dissolved in water to make up 500 cm3 of solution. (Ans:73.0 g dm-3 ) (b) the molarity of a solution which is prepared by dissolving 0.30 mol of sodium hydroxide,

NaOH in distilled water to make up 250 cm3 of solution. (Ans: 1.20 mol dm-3)Solution:

(a) (b) Question 2The molarity of a bottle of nitric acid, HNO3 solution is 2.0 mol dm-3. What is the concentration of sodium in g dm-3

[Relative atomic masses: H, 1; N, 14; O, 16] Ans: 126 g dm-3Solution;

Question 3

Calculate the molarity of a sodium sulphate, Na2SO4 solution with a concentration of 28.4 g dm-3. [Relative atomic masses: O, 16; Na, 23; S, 32] Ans: 0.2 mol dm-3Question 4

A student pipettes 25.0 mol dm3 of sodium hydroxide, NaOH solution into a conical flask. The concentration of the alkali was 1.5 mol dm-3. Calculate the number of moles of sodium hydroxide, NaOH in the flask. Ans: 0.0375 molPreparation of Standard Solutions by Dilution MethodStandard solution is a solution in which its concentration is accurately known.

Use this formula ok..

M1 V1 = M2 V2 M1 = molarity of the solution before water is added

V1 = volume of the solution before water is added

M2= molarity of the solution after water is added

V2= volume of the solution after water is added

a) Preparation of standard solution

- prepared by using a volumetric flask b) Preparation of a solution by dilution method

- adding distilled water to a concentrated solution

- changes the concentration of the solution- does not change the amount of soluteExample 1:Find the volume of 2.0 mol dm3 sulphuric acid, H2SO4 needed to prepare 100 cm3 of 1.0 mol dm-3 sulphuric acid, H2SO4.

Solution;

M1 V1 = M2 V2 2.0 mol dm-3 V1 = 1.0 mol dm-3 100 cm3 V1 = 1.0 mol dm-3 100 cm3 = 50 cm3 2.0 mol dm-3 Example 2:

Volume of 2.0 mol dm(3 nitric acid needed to be diluted with distilled water to make 250 cm3 of

0.5 mol dm(3 nitric acid?

Solution: Example 3:

50 cm3 of water is added to 200 cm3 of a 2 mol dm(3 solution of sodium hydroxide.

Determine the molarity of the diluted solution.

NeutralizationDefinition: The reaction between an acid and a base to produce a salt and water only. Example : i. HCl + NaOH NaCl + H2O

ii. H2SO4 + CuO CuSO4 + H2O The ionic equation for neutralization;

H+ + OH- H2O

Acid-Base Titration

IndicatorColour in alkalisColour in neutralColour in acids

Methyl orangeYellow OrangeRed

PhenolphthaleinPinkColourlessColourless

LitmusBluePurpleRed

Calculation involving neutralization using balanced equations.

Question 1: A student pipettes 25.0 mol dm-3 of sodium hydroxide, NaOH solution into a conical flask and filled a burette with 0.10 mol dm-3 hydrochloric acid, HCl to carry out titration. He obtained an average volume of 22.0 cm3 hydrochloric acid, HCl. What was the molarity of the sodium hydroxide, NaOH solution? [ans: 0.088 mol dm-3]Question 2:

What is the volume of 0.5 mol dm-3 sulphuric acid, H2SO4 needed to neutralize 25.0 cm3 of 0.8 mol dm-3 ammmonia, NH3 solution?[Ans: 20.0 cm3]Question 3:

A sample of copper(II) oxide, CuO was found to completely neutralize 100 cm3 of 0.5 mol dm-3 hydrochloric acid. Calculate the mass of sample. [Ans : 2.0 g]

Alkalis: only bases that soluble in water

Bases: all metal oxides or metal hydroxides

Switch

Switch

Glacial Ethanoic acid

Carbon

Carbon

Light bulb

Aqueous

Ethanoic acid

Carbon

Carbon

x molar mass (g mol-1)

g dm -3

Increasingly alkaline

Increasingly acidic

Neutral

pH

Ammonia in water

14

13

12

Bases and alkalis only show its alkalis properties when dissolved in water.

Ammonia

in tetrachloromethane

NH3 (l) NH4+ (aq) + OH (aq)

Ammonia ammonium ion hydroxide ion

+ H2O (l)

Light bulb

mol dm-3

MOLARITY

An acid only shows its acidic properties when dissolved in water.

Switch

Switch

11

Carbon

Carbon

Light bulb

10

Carbon

Carbon

Light bulb

9

8

7

6

5

4

3

2

1

0

Concentration

(mol /dm3)

Number of mole of solute (mol)

=

Volume of solution (dm3)

Concentration

(g /dm3)

Mass of solute/substance (g)

=

Volume of solution (dm3)

Retort stand

Burette

Hydrochloric acid, HCl

Conical flask

25 cm3 sodium hydroxide, NaOH + phenolphthalein

MA VA

MB VB

=

a

b

MA = Molarity of alkaliMB = Molarity of base

VA = Volume of alkaliVB = Volume of base

a = number of mole of acid from balanced chemical equation

b = number of mole of base from balanced chemical equation

Calibration mark

molar mass (g mol-1)

1

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